To get an idea of how long MergeSort takes, we count
the number of comparisons it makes in the worst case.
Call this function
M(n). We parameterize it by
the size of the array,
n, because MergeSort takes longer on longer inputs.

It is difficult to describe
M(n)
exactly, so instead we describe a simpler function
T(n)
which bounds
M(n)
from above,
i.e.
M(n) <= T(n).

Because MergeSort has two cases, the description of T also
has two cases. The base case is just

T(n) = 0, if n<2.

The induction case says that the number of comparisons
used to sort n items is at most
the sum of the worst-case number of comparisons for each
of the three steps of the induction case of MergeSort.
That is,

T(n) = T(n/2) + T(n/2) + n, if n>1.

Let's look at this expression one term at a time.

The first term accounts for the number of comparisons
used to sort the left half of the array. The left half of the array
has half as many elements as the whole array, so
T(n/2) is enough to account for all these comparisons.

The second term is a bound on the number of comparisons used
to sort the right half of the array. Like the left
half,
T(n/2) is enough here.

The last term, n is an upper bound on the number of comparisons
used to merge two sorted arrays. (Actually, n-1 is a tighter
bound, but let's keep things simple.)

Now, something interesting is going on here.
The inductive case
defines T in terms
of itself!
This is called a recursive definition.
The function T has a recursive definition because the
description of MergeSort itself is recursive.

When a numerical expression is defined recursively, we call its
set of defining equations a recurrence relation.

Recursive definitions (and recurrence relations)
only make sense if we ``hit bottom'' at some point.
That is, unless the recursion terminates, we can't assign meaning
to T(n). That's why we needed the base case.

In the beginning, there is an array of bars that are unsorted.
Unsorted bars are coloured blue (a dark grey on a black and white screen).

To sort the left half of the array, the program calls itself recursively,
passing the left
half of the bars down to a new activation of MergeSort. This
happens repeatedly until we have an array of only one element. As noted
earlier, one-element arrays are already sorted. Bars in sorted arrays are
coloured green (or light grey on a black and white screen).

When the left half of an array is sorted, we sort the right half.

Once the two halves of an array have been sorted by recursive calls,
MergeSort merges the two sorted halves into a sorted whole.
This is done by examining and comparing the smallest remaining bar in each
of the sorted halves. The two bars being compared are coloured
red (or medium grey). The smaller of these two bars turns green and
is moved up and
tacked onto the right end of the parent array.

When one of the two subarrays becomes empty,
the remaining bars in the other subarray are copied
in order into the parent array. No comparisons need to be made.

That describes the workings of MergeSort. I suggest you go back, click on
the ``Load input'' button, then click on the ``Go'' button, and adjust the
speed slider so that you can see the comparisons being made. Then come back.

But what about those numbers over the bars?
Each array of bars (except for those with only one bar)
has an inequality associated with it.
The right hand side of the inequality is the upper bound on the
number of comparisons required to sort an array of that size.
That is, the right hand side is T(n),
where n is the number of bars in that array.
The left hand side is the actual number of comparisons made so far.
This latter figure gets updated as the sorting proceeds.

I have used strict inequality because we have actually overestimated
the number of comparisons required for merging by 1 at each level.
(To tell you the truth,
it's really because I couldn't find a nice-looking less-than-or-equal-to sign
that would fit in that space!)

You can specify various canned inputs, i.e. inputs that I've programmed
for you, by bringing up a pop-up menu where the word ``Strange'' originally
appeared.
Note that once you have specified an input, you have to click on ``Load input''
in order for it to be loaded into the array for sorting.
Try the various options.

If you are adventurous, you can type in your own values for the heights of the
bars by first clicking on the text box next to the input choice
menu, and then typing. Actually, you can specify any set of numbers you
like, but only the first 16 will be looked at.
Even then, the program doesn't use the absolute values as bar heights, but only
their relative ordering. If you type in fewer than 16 numbers, then the
last number is repeated. For example, to see what MergeSort does
when when all the inputs are the same, type in just a single 1 in the
text box, then load the input, and click on ``Go''.

If you are adventurous but lazy, then you can just click on the button
labelled ``Shuffle now''. This randomly rearranges the numbers in the text
box. For interesting results, select the ``Ascending'' input and then
shuffle it a few times.

To better understand all this, try the following exercises:
(Note to 238 students: these are not required for your course.
I just made these up.)

What if n isn't a power of 2?

Tightening the estimation:

Write down a recurrence relation for T'(n) which uses the
tighter estimate of n-1 comparisons for merging two lists into one
list of length n.

Compute the value of T'(16).

Can you find a 16-element input that forces MergeSort to use T'(16)
comparisons?

Can you describe a worst-case input for MergeSort for general n?

Can you find a closed form solution for T(n)?

Can you find a closed form solution for T(n) to within multiplicative
and additive constants, i.e. find a closed form expression f(n) so that
T(n) is O(f(n))? Hint: the answer to that is on this
web page, but its derivation is not.

What, besides comparisons, takes time when actually sorting arrays?

What resources, besides time, does the above implementation of MergeSort
use? How much? Can you make MergeSort more efficient in that respect?